The category $bf{FinVect}$ of finite vector spaces is rigid.
$begingroup$
I am following Pavel Etingof et al's book on tensor categories.
They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".
My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.
category-theory monoidal-categories
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
$endgroup$
add a comment |
$begingroup$
I am following Pavel Etingof et al's book on tensor categories.
They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".
My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.
category-theory monoidal-categories
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
$endgroup$
1
$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33
$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14
add a comment |
$begingroup$
I am following Pavel Etingof et al's book on tensor categories.
They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".
My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.
category-theory monoidal-categories
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
$endgroup$
I am following Pavel Etingof et al's book on tensor categories.
They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".
My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.
category-theory monoidal-categories
category-theory monoidal-categories
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
edited Dec 4 '18 at 13:33
Soap
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
asked Dec 4 '18 at 11:27
SoapSoap
1,022615
1,022615
1
$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33
$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14
add a comment |
1
$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33
$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14
1
1
$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33
$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33
$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14
$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14
add a comment |
1 Answer
1
active
oldest
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$begingroup$
What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$
where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.
This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.
Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$
answered Dec 4 '18 at 11:44
NephryNephry
514512
$endgroup$
6
$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52
add a comment |
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1 Answer
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$begingroup$
What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$
where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.
This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.
Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$
answered Dec 4 '18 at 11:44
NephryNephry
514512
$endgroup$
6
$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52
add a comment |
$begingroup$
What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$
where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.
This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.
Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$
answered Dec 4 '18 at 11:44
NephryNephry
514512
$endgroup$
6
$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52
add a comment |
$begingroup$
What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$
where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.
This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.
Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$
answered Dec 4 '18 at 11:44
NephryNephry
514512
$endgroup$
What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$
where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.
This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.
Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$
answered Dec 4 '18 at 11:44
NephryNephry
514512
answered Dec 4 '18 at 11:44
NephryNephry
514512
answered Dec 4 '18 at 11:44
NephryNephry
514512
answered Dec 4 '18 at 11:44
NephryNephry
514512
514512
6
$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52
add a comment |
6
$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52
6
6
$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52
$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52
add a comment |
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$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33
$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14